Question

Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award x each, y each and z each for the three respective values to $3,2$ and $1$ students respectively with a total award money of $1,600$. School B wants to spend $2,300$ to award its $4,1$ and $3$ students on the respective values (by giving the same award money to the three values as before). If the total amount for one prize on each value is $900$, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award

Answer 1

Let matrix C represents number of students receiving prize for the three categories :

Let $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],$ where $x,y\xi z$ are rupees mentioned as it is the question, for sincerity, truthfulness and helpfulness respectively.

Let, $D=\left[\begin{array}{c}1600\\ 2300\\ 900\end{array}\right]$ is a matrix representing total award money for school $A,B$ and for one prize for each value.

We can represent the given question in matrix multiplication as :

$\Rightarrow CX=D$

or $\left[\begin{array}{ccc}3& 2& 1\\ 4& 1& 3\\ 1& 1& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1600\\ 2300\\ 900\end{array}\right]$

Now, solution of the matrix equation exist if $101\ne 0$

i.e, $\left|\begin{array}{ccc}3& 2& 1\\ 4& 1& 3\\ 1& 1& 1\end{array}\right|=3(1-3)-2(4-3)+1(4-1)$

$=-6-2+3$

$=-5$

Therefore, the solution of the matrix equation is $X=C^{-1}D$

$\Rightarrow C^{-1}=\frac{1}{\left|C\right|}adj\left|C\right|$

Co-factor matrix of $C=\left[\begin{array}{ccc}-2& -1& 3\\ -1& 2& -1\\ 5& -5& -5\end{array}\right]$

$\Rightarrow adj\left(C\right)=\left[\begin{array}{ccc}-2& -1& 3\\ -1& 2& -1\\ 5& -5& -5\end{array}\right]$

$\Rightarrow adj\left(C\right)=\left[\begin{array}{ccc}-2& -1& 5\\ -1& 2& -5\\ 3& -1& -5\end{array}\right]$

$\therefore C^{-1}=\frac{1}{-5}\left[\begin{array}{ccc}-2& -1& 5\\ -1& 2& -5\\ 3& -1& -5\end{array}\right]$

Now, $X=C^{-1}D$

$=\frac{1}{-5}\left[\begin{array}{ccc}-2& -1& 5\\ -1& 2& -5\\ 3& -1& -5\end{array}\right]\left[\begin{array}{c}1600\\ 2300\\ 900\end{array}\right]$

$=\frac{1}{-5}\left[\begin{array}{c}-3200-2300+4500\\ -1600+4600-4500\\ 4800-2300-4500\end{array}\right]$

$=\frac{1}{-5}\left[\begin{array}{c}-1000\\ -1500\\ -2000\end{array}\right]=\left[\begin{array}{c}200\\ 300\\ 400\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}200\\ 300\\ 400\end{array}\right]$

$\therefore x=200,y=300,z=400$

Award can also given for punctuality.

Hence, the answer is $x=200,y=300,z=400.$